Question

我有一个10x20 numpy数组

[[255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255],
 [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
 [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
 [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
 [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
 [255,255,255,255,255,255,255,0,0,0,255,255,0,0,255,255,255,255,255,255],
 [255,255,255,255,255,255,255,0,0,0,255,255,0,0,255,255,255,255,255,255],
 [255,255,255,255,255,255,255,0,0,0,255,255,0,0,255,255,255,255,255,255],
 [255,255,255,255,255,255,255,0,0,0,255,255,0,0,255,255,255,255,255,255],
 [255,255,255,255,255,255,255,0,0,0,255,255,0,0,255,255,255,255,255,255]]

我希望以一系列维度的形式重复数字0（0和39的岛）。如上例所示，如果我将数字0作为参数传递，则函数应从上到下穿过矩阵，然后从左到右，从而产生维数组。以上示例的输出应为

#Zeroes repetition dimensions
[[20,4],[3,5],[2,5]]

如何获得此输出。任何numpy函数来做到这一点？提前谢谢。

Answer 1

我会使用itertools groupby：

import numpy as np
from itertools import groupby
from collections import Counter
arr=np.array([[255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255],
 [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
 [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
 [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
 [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
 [255,255,255,255,255,255,255,0,0,0,255,255,0,0,255,255,255,255,255,255],
 [255,255,255,255,255,255,255,0,0,0,255,255,0,0,255,255,255,255,255,255],
 [255,255,255,255,255,255,255,0,0,0,255,255,0,0,255,255,255,255,255,255],
 [255,255,255,255,255,255,255,0,0,0,255,255,0,0,255,255,255,255,255,255],
 [255,255,255,255,255,255,255,0,0,0,255,255,0,0,255,255,255,255,255,255]])

val_to_check=0
agg_res=[] 
for a in arr:
    groups = groupby(a)
    temp = [[label, sum(1 for _ in group)] for label, group in groups]
    for t in temp:
        if t[0]==val_to_check:
            agg_res.append(t)

res_to_group=[l[1] for l in agg_res]
final_res=Counter(res_to_group)
#or, in list form: 
res_list=map(list,final_res.items())

输出：

[[2, 5], [3, 5], [20, 4]]

或者，如果你想要，作为一个功能：

def find_islands(arr,val):
    agg_res=[] 
    for a in arr:
        groups = groupby(a)
        temp = [[label, sum(1 for _ in group)] for label, group in groups]
        for t in temp:
            if t[0]==val:
                agg_res.append(t)

    res_to_group=[l[1] for l in agg_res]
    final_res=Counter(res_to_group)
    #or, in list form: 
    return map(list,final_res.items())

示例运行：

In[65]: find_islands(arr,255)
Out[65]: [[2, 5], [20, 1], [6, 5], [7, 5]]

在numpy中以维度的形式重复数字

1 个答案: